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#1
August 5th, 2003, 02:03 PM
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Help with simple XSL
I have an XML document, which describes university style guides for web pages. The structure to the document looks like this.
<styleguide> <rule> <name>style guide rule</name> <description>Some content <link href=”some url”>content that is a hyperlink </link> more content</description> </rule> <rule> <name>another style guide rule</name> <description>Content <link href="some url">content that is a hyperlink </link> More content </description> </rule> </styleguide> My question is how can I retrieve the link information and transform it to a HTML hyperlink? Not all description tags have content that are links but some do. I have read a number of Xpath and XSLT tutorials because I am new to the XML world but I just can’t figure out how to deal with the link element if it exisists within a description element. Any help would be greatly appreciated. Thanks!
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#2
August 5th, 2003, 03:34 PM
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Try:
Code:
<xsl:if test="description/link">
<a href='{description/link/@href}'>
<xsl:value-of select="description/link/text()">
</a>
</xsl:if>
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#3
August 5th, 2003, 04:06 PM
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Thanks for the reply. I guess I need to be a little more clear. If my description has a link element in the middle of the content I can get the first part of the content to display and the link element content, but I can't get the rest of the description content to display.
For example: <description>this is text in my description. <link url="some url">This text is a hyperlink</link> this is more text in my description.</description> After I run into a link element I can't get the rest of the content in my description tag to appear. Thanks again for the reply. Any help on this would be great.
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#4
August 5th, 2003, 05:15 PM
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apply this style sheet against your xml and see if it gives you a starting point. It will work for your xml data as long as it only has one <link> per description. You probably will be able to expand the scale with some time:
Code:
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:template match="/">
<xsl:for-each select="styleguide/rule">
<xsl:value-of select="concat('Name: ', name)" />
<br />
<xsl:for-each select="description">
<xsl:value-of select="substring-before(., link)" />
</xsl:for-each>
<xsl:for-each select="description/link">
<a href="http://{@href}">
<xsl:value-of select="." />
</a>
</xsl:for-each>
<xsl:for-each select="description">
<xsl:value-of select="substring-after(., link)" />
</xsl:for-each>
<br />
</xsl:for-each>
</xsl:template>
</xsl:stylesheet>
Let me know if it helps...
__________________
mr... mike.rusaw@realpage.com RalPage, Inc. "I have made this letter longer than usual, only because I have not had the time to make it shorter." - Blaise Paschal
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#5
August 6th, 2003, 10:59 AM
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I just ran the style sheet and it works great. I think this will be a good enough starting point for me to figure out how to deal with multiple links are zero links. Thanks for the help and the reply.
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